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Proof step 6 from proof fillsgn-rec (previous step)

¬(# bytes < m + 1 ∧ bytes2int(bytes) < 0), # bytes ≤ m ⊦ fillsgn(m + 1, bytes) = (¬ 0 ≤ b2i(bytes.first) ⊃ i2b(~1);i2b(0)) ' + fillsgn(m, bytes), bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bytes ≤ m, 0 ≤ bytes2int(bytes) ⊦ fillsgn(m + 1, bytes) = (¬ 0 ≤ b2i(bytes.first) ⊃ i2b(~1);i2b(0)) ' + fillsgn(m, bytes), bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ n → (n < m ↔ false);% ⊦ m < n + 1 ↔ ¬ n < m;